PHYSICAL CHEMISTRY II

CHAPTER 9: Chemical Kinetics I

Rate of Consumption, Formation, and Reaction

Consider the reaction . The numbers of moles of the species at the beginning of the reaction (at time ) are , respectively.

Assuming the reaction thereafter proceeds from left to right, the moles at some time become , respectively, where is the extent of the reaction in terms of moles.

Now, we can define the moles of reactants and product after time as:, such that are constants and are functions of the reaction extent, i.e. .

Furthermore, we may derive these expressions with respect to time and obtain:

Solving for gives the rate of reaction in terms of moles:a

However, it is more typical to track concentrations than moles. So we may modify the above expression as follows.

Let be the volume of the reaction chamber (assuming it remains constant throughout the reaction). Then:

If we define the extent of the reaction in terms of concentration as and the concentrations of the reactants and product as respectively, then the rate of reaction in terms of concentration is:

Specifically, we can define the rate of reaction in terms of concentration with respect to each species:

Thus: .

Hence, we can reconsider the reaction in terms of concentration such that:

The reaction can also be represented graphically:

Table 1: Rates w.r.t species at t = 0 and t = any.Initial rates (at time )Rates at any time ()

Table 2: Summary of rate expressions.Moles at

Moles at

Moles derived w.r.t time

Reaction rate in moles

Concentration at

Concentration at

Concentration derived w.r.t time

Reaction rate in concentration

Empirical Rate Equations

For the reaction , the rate of consumption of A can be expressed empirically by an expression of the form , called an empirical expression. We may change this expression to an equation by adding a proportionality constant k: , where kA, , and are independent of concentration and of time.

Similarly, for product Z, where kZ is not necessarily the same as kA, the rate of formation of product is .

When these equations apply, the rate of reaction must also be given by an equation of the same form: .

Thus, (the rate law in differential form).If we let , then . Likewise, . If we let , then .

In these equations, kA, kZ, and k are not necessarily the same, being related by stoichiometric coefficients; thus if the stoichiometric equation is , then .

Order of Reaction

The exponent in the previous equations is known as the order of reaction with respect to A and can be referred to as a partial order. Similarly, the partial order is the order with respect to B. These orders are purely experimental quantities and are not necessarily integral; that is, they are independent of stoichiometry. The sum of all the partial orders, , is referred to as the overall order and is usually given the symbol n.

A first order reaction is one in which the rate is proportional to the first power of the concentration of a single reactant, such that . For example, the conversion of cyclopropane to propylene is 1st order.

In a second order reaction, the rate must be proportional to the product of two concentrations [A] and [B] if the reaction simply involves collisions between A and B molecules. For instance, in the reaction , the rate from left to right is proportional to the product of the concentrations of the two reactants. That is, , where k1 is a constant at a given temperature. The reaction from left to right is said to be 1st order in H2, 1st order in I2, and 2nd order overall. The reverse reaction is also 2nd order; the rate from right to left is proportional to the square of the concentration of HI: .

Similarly, the kinetics must be third order if a reaction proceeds in one stage and involves collisions between three molecules, A, B, and C.

Examples

For the reaction , . Assuming that = 1 and = 1, the forward reaction is 2nd order and .

For the reaction , .Assuming that = 2, the forward reaction is 2nd order and .

For the reaction , .Assuming , the forward reaction is of order and .

For the reaction , .Assuming that = 1, the forward reaction is 1st order and .Also, and so .Integrating gives , where C is a constant. This is the rate law in integral form.At and , .Therefore, , and , and .Get rid of natural log: . Hence, is the rate law for the disappearance of A.At , .At , .In other words, for the reaction :The initial concentrations of A and B are [A]0 and 0, respectively.The concentrations after time are and .Hence and . is the rate law of appearance of B.

Integration of Rate Laws

First Order Reaction:

Consider .At , and .At , , where a, ax, and [A]0 are constants.

The derivative of [A] with respect to time gives us . Therefore, , where is the partial order of reaction with respect to A, determined experimentally. Assuming the reaction is 1st order, . Thus, .

To simplify, we can let and let .

Rearranging therefore gives , the rate law in differential form.

Important Integrals

(1)

(2)

(3)

(4)

(5)

(6)

If we integrate the above equation, we get the following:, the rate law in integral form.

When , . Therefore, .

Substituting gives: . Rearranging, and .

We now define the half-life of the reaction to be the time required for [A] to drop to half its value; that is, .

Thus, for a 1st order reaction, at t:

is independent of concentration in 1st order reactions.

Second Order Reaction:

Consider .

, where is the partial order of reaction with respect to A, determined experimentally. Assuming the reaction is 2nd order, . Thus, .

To simplify, we can let and let .

Rearranging therefore gives , the rate law in differential form.

If we integrate the above equation, we get the following:, the rate law in integral form.

When , . Therefore, .

Substituting gives: . Rearranging, .

Thus, for a 2nd order reaction, at t:

depends on in 2nd order reactions.

Third Order Reaction:

Consider .

, where is the partial order of reaction with respect to A, determined experimentally. Assuming the reaction is 3rd order, . Thus, .

To simplify, we can let and let .

Rearranging therefore gives , the rate law in differential form.

If we integrate the above equation, we get the following:, the rate law in integral form.

When , . Therefore, .

Substituting gives: . Rearranging, .

Thus, for a 3rd order reaction, at t:

depends on in 3rd order reactions.

nth Order Reaction:

Consider .

, where is the partial order of reaction with respect to A, determined experimentally. Assuming the reaction is nth order, . Thus, .

To simplify, we can let and let .

Rearranging therefore gives , the rate law in differential form.

If we integrate the above equation, we get the following:, the rate law in integral form.

When , . Therefore, .

Substituting gives: . Rearranging, .

Thus, for a nth order reaction, at t:

We can invert the initial concentration term to send it to the number with the index , flip the fraction to also give it the index , and divide both sides of the equation by (to isolate t and distribute the negative sign to the denominator, ). Thus:

depends on in nth order reactions.

Example:

Consider the 1st order reaction . At time , and .Hence, and .If , then or .Therefore, .Also, is a separable equation such that .Integrating gives , where C is an integration constant.At time : and .Then .And if , then .In terms of [B]: .This is a separable equation such that: .Integrating gives .At time : and .Then , as above.

Example

Consider the 2nd order reaction or .Let . And let .Then rearranging gives .Integrating gives .At time : .Then , where k has units .At time : .

Consider a 2nd order reaction that has two reactants, such as .We can express the rate as , but it cannot be integrated as it contains two different variables.We still know that at time : , , and .Similarly, at time :, , and .If we assume that , then as well.The rate then becomes , a separable equation.Thus, , which can be integrated as in the example previous example.

Consider a 3rd order reaction .The rate is . Let .Then rearranging gives .Integrating gives .At time : .Then .At : GO OVER TO CLARIFY!

Example

Consider the reaction or .Then , where and at any time t.If we let , then .Then .And since , then (and thus and ).Therefore, .

Example

Consider a reaction .At time , we assume that , , and .Likewise, at time , , , and .The rate is given by: .Integrating gives .At , . Therefore, .And at , .HAVE HOKMABADI CHECK THIS!

Example

Consider the 2nd order reaction .At time , we assume that and .Similarly, at time , and .The rate is then given by: .We will denote and .Then .We use partial fractions to solve for a and b.

Let

We can show that and that .

Integrating (integral #6) gives: .

Review

At time : and , so that .Thus, .Using the integral , we get:

Or .At , .Therefore, .Multiplying both sides by gives Rearranging gives Thus And Hence And WAIT FOR CORRECTION

Reversible 1st Order Reactions

Consider the reaction: .This overall reaction is composed of two elementary reactions: for which the rate is: ; and for which the rate is: The overall rate is .Before we can integrate, we must express B in terms of A.

Hence Rearranging gives:

Integrating (integral #6) gives: .At :

Hence

At equilibrium:

and since [A] is a constant.

Hence .

Therefore, , which is in accordance with the equilibrium condition that the forward rate should equal the reverse rate.

Rearranging this gives , which we will name the equilibrium rate constant, a chemical thermodynamics term.

We solve for and by substitution.

Rearrange (2): (3) Substitute the above into (1): Factor out : Make k-1 the common denominator: Make the subject: Note that Substitute into (3): The k-1 term cancels to give: .

To prove these solutions for and hold, we substitute them into (2) and cancel common terms:

Furthermore, we can substitute our new expression for into the original integral equation, :

The term is common to both numerator terms as well as the denominator and can be canceled out; k-1 in the numerator can also be canceled out. This gives:

Show that: .

These two expressions for must be equivalent:

Review

Consider the reaction: .

This overall reaction is composed of two elementary reactions: and .

The overall rate in differential form is: .

The integral form is: . (Eq. 1)

This simplifies to give: . (Eq. 2)

We can show that:

1. :

At time : Make k-1 the common denominator: The terms and both cancel out in the numerator and denominator to give: at time , which is true.

2. :

At time : Therefore: This boils down to at time , which is true.

Hence both equations are valid.

Relative rate constants in a reversible 1st order reaction

Consider the reaction , composed of two elementary reactions: and .

If we suppose that , then the forward reaction predominates and the integrated rate law for the overall reaction (Eq. 1) changes as follows:

The equation for the concentration of product B changes as follows:

Hence we can write a new expression for :

If, similarly, we suppose that ; that is, that , then and Eq. 2 becomes:

Dynamic Equilibrium in a Reversible 1st Order Reaction

When a reversible 1st order reaction reaches dynamic equilibrium, the following conditions are true:

Consecutive 1st Order Reactions

Consider the reaction.

At time , , , and .

At time , , , and .

As we define , we can say that

The rates with respect to each species are as follows:, , and .

With some manipulation, we can formulate equations for the concentrations of A, B, and C.

(1)(2)(3)

We can substitute into the expression for as follows: (4)

We can also differentiate the expression with respect to t by using the product rule and then factor out the common term, :

The term in parentheses above is equivalent to (4) and so we can substitute and use the laws of exponents to simplify:

Now we can rearrange this equation and integrate using :

At time , the equation becomes: .

Hence, at time :.

Factoring the common term , rearranging, and simplifying gives an expression for [B]:

Finally, as we have stated that , we can solve for [C] as follows:Rearrange to make [C] the subject:

Substitute equations for [A] and [B]:

Factor :

FINAL ANSWER SHOULD BE

The Rate-Determining Step (RDS)Suppose now that .This would mean that B forms C much more quickly than A forms B. Therefore, remains small and almost constant throughout the reaction. That is, whenever a molecule of B is formed, it decays rapidly into C.Hence the equation for becomes .

We say that is very small and remains constant over the course of the reaction.

Similarly, becomes , and .

All three of these equations depend only on k1, thus proving that the 1st step determines the rate (or is the rate-determining step) of the reaction.

We say that a steady state approximation holds for species B, since [B] is constant and small over the course of the reaction.Therefore, the rate with respect to B is given by: .Hence and .

Similarly, the rate with respect to C is: .Substituting the final expression above for B gives: Finally, we substitute for [A] to obtain: .This is an equation in [A] only and can be integrated to provide a new equation for [C] as follows.Rearrange: Integrate using : Simplify: At time , therefore: .Thus: Or: , the same expression attained via the steady state approximation.

Reaction Mechanisms

Consider the reaction .

The nitrogen monoxide-catalyzed oxidation of gas gives the overall stoichiometry of the reaction but does not tell us the process, or mechanism, by which the reaction actually occurs. This reaction has been postulated to occur by the following two-step process:

The number of times a given step in the mechanism occurs for each occurrence of the overall reaction is called the stoichiometric number S of the step.

Do not confuse the stoichiometric number S of a mechanistic step with the stoichiometric coefficient of a chemical species.

Example

The gas-phase decomposition of has overall reaction:

Species like and , which do not appear in the overall reaction, are called reaction intermediates. Each step in the mechanism is called an elementary reaction. A simple reaction consists of a single elementary step. A complex/composite reaction consists of two or more elementary steps.

The Diels-Alder addition of ethylene to butan-1,3-diene to give cyclohexene is believed to be simple, occurring as a simple step:

Rate-Determining Step Determination

Consider the composite reaction .

If we assume that , then the overall reaction becomes .

The concentration of B, , remains very small and constant during the course of the reaction, such that . Hence, and .We also know that . Thus, .Finally, .

This shows that the rate of appearance of A depends on the 1st elementary step, which by virtue of the large size of is assumed to be the slower, rate-determining step.

Example

Consider the reaction .We define: .

We know that:

Therefore:

The equation becomes and can be integrated to give: .Thus, .Although the reaction is , the overall reaction is , due to the steady state approximation. Hence:

Example

Consider the reaction .

We know that:

Therefore:

Thus:

If we let and , then:

The reaction can therefore be simplified to .And if , then .

Furthermore, if , then . Likewise, , then .

Example

Kinetics of Relaxation

Consider the reaction .

We know that:

Hence, the rate is given by .

At equilibrium: .

If is the concentration of Z at equilibrium (i.e. if ), then:

We define the change in concentration of x as: . Thus:

This is the rate law in differential form.

Rearranging gives .

Integrating gives: .

At time : . Therefore:

This is the rate in the form , where , , and .

A graph of would therefore look like:

A plot of , however, would give a completely different graph:

The Principle of the Temperature Jump Technique

The relaxation time is defined as the time corresponding to and . Therefore, becomes .And so: and .

Thus, the negative reciprocal of the slope of a graph of is equal to the relaxation time .

N.B The equilibrium constant is given by .

Example

Consider the reaction .

Let , , and .The rate is .

If we let , then:

At equilibrium, .Therefore, .

If , then:

Terms only in cancel out.

If the displacement from equilibrium is only slight, then , so that may be written as .

Rearranging gives .

Integrating gives .

At time , and therefore .

Thus:

The relaxation time is defined as that corresponding to .

The dissociation of a weak acid, , can be represented as .

The rate constants k1 and k-1 cannot be determined by conventional methods but can be determined by the T-jump technique.

We can prove that the relaxation time is given by , where xe is the concentration of the ions (Y and Z) at equilibrium.ASK HOKMABADI FOR THE HW BACK IN ORDER TO SEE THE ANSWER TO THIS.

The Bohr Hydrogen or Hydrogen-like (isoelectronic to hydrogen) Atom

Consider a hydrogen bulb shining light upon a concave lens in front of a triangular pris. If a screen is placed behind the prism, discrete bands of colored and invisible light (see below).

As the capacitor plates in a hydrogen bulb accumulate charge, the negatively charged plate emits a beam of electrons that bombards hydrogen atoms, which in turn release photons of light as they are excited.

Compare hydrogen to some hydrogen-like atoms:

H1 proton1 electron

He+2 protons1 electron

Li2+3 protons1 electron

1 electron

Let , , , and .

1. The attractive force between negative and positive charges (protons and electrons) is the normal type of Coulombic attraction, such that: , where , is the permittivity of the vacuum. Therefore:

2. The electron can remain in any particular state (fixed radius) indefinitely without radiating its energy, provided that the electrons angular momentum is (Equation 20.10), where (quantum number, , Plancks constant, , , , ).Hence , and possible values of L are: , , , , Also, L has the same units as h: joule per second.

Note: The total energy of the atom is equal to the sum of the kinetic energy of orbiting electrons and the potential energy of electrons attracted to protons.Mechanical Stability

Mechanical stability requires that the attractive and centripetal forces cancel each other; that is:(Eq. 20.13)

From the quantization of Eq. 20.10 , we can write that: (Eq. 20.15)where and .

Substitute for v2 in Eq. 20.13:

Therefore: , where , .

Figure 1: Proof of a0.

This proves that the distance of the electron from the nucleus is quantized.

Note that for the hydrogen atom, and thus .

Proof that kinetic energy (T), potential energy (U), and total energy are also quantized:

The force is given by:

Integrating gives:

When , and .

Then .

The kinetic energy is is given by: .

Substitute Eq. 20.13: . That is: .

Therefore: .

We see that Etot is negative and inversely proportional to r.

The final expression for the (total) energy is obtained by substituting the value of r as expressed in Eq. 20.17 below:

and

where and

Therefore:

That is: .(Eq. 20.17)where .

The Bohr Theory/Model of the Hydrogen Atom

Let .

Let , , and , where , , and .

The potential energy is .The kinetic energy is .The total energy is

The energy required to move an electron from to is called its ionization energy (IE).

The term is called the threshold frequency.

Hydrogen-like Atoms

Angular momentum is an integral multiple of , such that:, where

From this, we can get that:

i.

ii.

iii.

iv. , since this change in energy is due to absorption or emission of photons. For a hydrogen atom, . Therefore: .Therefore: .And

If we let and , then: . This is the Lyman series (UV).If we let and , then: . This is the Balmer series (visible).If we let and , then: . This is the Paschen series (IR).

De Broglie Wavelength

The De Broglie wavelength is given by: , where .

Since , then ; that is, .

ExampleIf , then .

In the diagram above, .

This comes from the standing wave shown below:

Joining the ends of the string, a and b, creates the pattern and circle above.

When , .Therefore, .

When and , .When and , .

Now, , where and .If , then .Since , then:

And so:

when and .

Generally:

When and , then .When and , then .When and , then .

Therefore, we can say that:

Then we can find the speed for other atoms:

for H for He2+ for Lr102+ For what value of z does v1 match the speed of light?

No element with atomic number (z) 13697.

The Photoelectric Effect

Consider the horizontal, parallel, clean metal plates (see below) connected by a circuit containing an ammeter. Light of fixed intensity and varying frequency is shone upon the lower plate at an angle. Above a given threshold frequency, electrons jump from the lower to the upper plate, traveling through the circuit to generate a measurable current.

In other words, a metal with a clean surface is placed in vacuum and illuminated with light of a known frequency. If the frequency is greater than a particular minimum, or threshold frequency, electrons are instantly liberated fron the metal surface at a rate proportional to the light intensity (increasing the intensity does not liberate more electrons).

The energy is given by .

The total energy is conserved; therefore, where , , , , and .

Note that per electron.

Conservation of energy requires that:

where

and A plot of the frequency against the stopping voltage is shown below:

The equation of the graph is of the form .

Refinement of the Bohr Model

Consider the system below consisting of two masses the nucleus (with charge ze and mass mn) and the electron (with charge ze and mass me):

The moment of inertia gives rise to the relationship below:

The top equation rearranges to and . The bottom equation rearranges to and .

Then:

The moment of inertia about the center of mass is given by:

Then:

Expanding gives:

If we we define the reduced mass as , then:

And since , then , and we can approximate that:

Foundations of Quantum Mechanics

Two vectors and can be added or subtracted as follows:

A vector whose magnitudeis unity is called a unit vector.

SKIPPED SOME STUFF

Operators

An operator is a symbol that indicates that a particular operation is being performed on what follows the operator. For example, the square root operator is and can operate as follows: . The differential operator operates as in .

We denote operators in quantum mechanics by using the symbol hat, , as in the operator .

ExamplesIf and , then .Then And

ExampleIf , then

Reactions Having No Order

Not all reactions behave in the manner aforementioned, and the term order should not be used for those that do not. Instead, such reactions are usually enzyme-catalyzed and frequently follow a law of the form:

In the above equation, V and Km are constants, while [S] is a variable known as the substrate concentration. This equation does not correspond to a simple order, but under two limiting conditions, an order may be assigned:i. If the substrate concentration is sufficiently low, so that , then the law becomes , and the reaction is then 1st order with respect to S; orii. If [S] is sufficiently large, so that , then the law becomes , and the reaction is said to be 0th order (meaning the rate is independent of [S]).

Consider a reaction .Then .This is separable such that , the rate law in differential form.Integrating gives .At time : .Then , the rate law in integral form.At : , where k has units .

Rate Constants and Rate Coefficients

The constant k used in such rate equations is known as the rate constant or rate coefficient, depending on whether the reaction is believed to be elementary or to occur in more than one stage (respectively).

The units of k vary with the order of the reaction, as shown in the table below.OrderRate EquationUnits of kHalf-life, t

Differential FormIntegrated Form

0

1

2

2for reactants with different concentrations

n